Given two strings s and t, determine if they are both one edit distance apart.
Note:
There are 3 possiblities to satisify one edit distance apart:
Insert a character into s to get t
Delete a character from s to get t
Replace a character of s to get t
Example 1:
Input: s = "ab", t = "acb"
Output: true
Explanation: We can insert 'c' into s to get t.
Example 2:
Input: s = "cab", t = "ad"
Output: false
Explanation: We cannot get t from s by only one step.
Example 3:
Input: s = "1203", t = "1213"
Output: true
Explanation: We can replace '0' with '1' to get t.
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Approach #1 One pass
classSolution {publicbooleanisOneEditDistance(String s,String t) {int ns =s.length();int nt =t.length();if (nt < ns) {returnisOneEditDistance(t, s); }if (nt - ns >1) returnfalse;for (int i =0; i < ns; i++) {if (s.charAt(i) !=t.charAt(i)) {if (ns == nt) {returns.substring(i +1).equals(t.substring(i +1)); } else {returns.substring(i).equals(t.substring(i +1)); } } }return (ns +1== nt); }}
Approach #2 Two Pointers
publicbooleanisOneEditDistance(String s,String t) {int m =s.length(), n =t.length();if (Math.abs(m - n) >1) returnfalse;int k =Math.min(m, n);int i =0, j =0;while (i < k &&s.charAt(i) ==t.charAt(i)) ++i;while (j < k - i &&s.charAt(m -1- j) ==t.charAt(n -1- j)) ++j;return m + n - k -1== i + j;}